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https://designers-guide.org/forum/YaBB.pl Design >> RF Design >> PM component of a noisy signal in Oscillators https://designers-guide.org/forum/YaBB.pl?num=1459266167 Message started by iVenky on Mar 29th, 2016, 8:42am |
Title: PM component of a noisy signal in Oscillators Post by iVenky on Mar 29th, 2016, 8:42am Hi, I am reading the oscillator chapter in RF Microelectronics textbook by Razavi. I have one question that is bothering me for a long time that I am not able to completely understand. In the chapter on oscillator when he is talking about Conversion of Additive Noise to Phase noise, he considers x(t)=A cos(wot)+a cos(wo+Δw)t Where the second term is the additive noise component. He says that after passing the above signal through a limiter (so that amplitude noise component gets suppressed), we get the following FM (or PM) signal at the output of the limiter. xlim(t) = (A/2) cos wot - (a/2) cos(wo+Δw)t + (a/2) cos(wo-Δw)t I am able to understand the second and third terms of the above equation, but I don't understand how we are specifically getting (A/2) (in (A/2) coswot) . I mean why is it half of A? This has been bothering me a lot Thanks in advance |
Title: Re: PM component of a noisy signal in Oscillators Post by raja.cedt on Mar 30th, 2016, 12:03am HI- When you pass through limiter, ideally Am components will supressed because of hard clipping and phase noise will be passed with out having any attenuation. Now by using this you can split the about signal half as an AM and half as FM, hence out put only consists of A/2. Hope this helps, Raj. |
Title: Re: PM component of a noisy signal in Oscillators Post by iVenky on Mar 30th, 2016, 12:07am I understand that the phase noise component affects AM and PM equally (which is 'a' in this case). Why does 'A' also divide equally between AM and PM? Thanks for your help |
Title: Re: PM component of a noisy signal in Oscillators Post by raja.cedt on Mar 30th, 2016, 1:21am Hello- Please split the whole signal (not only a but also A, because for modulation you cant analyse signals independently) into AM and FM components. Best regards, Raj. |
Title: Re: PM component of a noisy signal in Oscillators Post by iVenky on Mar 30th, 2016, 3:31pm Hi, Thank you very much for the reply. I thought about this and felt that that should be the reason, but one thing that I don't understand is I am able to explain why 'a' divides equally between FM and AM, but I am not able to explain why 'A' divides equally. Could you please explain me your reason just to get another opinion? Thanks for your time! It really means a lot to me!!! :) |
Title: Re: PM component of a noisy signal in Oscillators Post by raja.cedt on Apr 3rd, 2016, 3:01am Hello- Any noise will split into PM and AM equally. But it is not correct to think like noise split but carrier won't, because how can you have spurs without carrier, they always be together otherwise you would call spur as another carrier rather spurs. That's I can tell.. Best Regards, Raj. |
Title: Re: PM component of a noisy signal in Oscillators Post by iVenky on Apr 3rd, 2016, 12:06pm Hi, Thanks for the reply. Could you please provide me the mathematical proof for that or some link or book where I can find that? Thanks a lot for your time |
Title: Re: PM component of a noisy signal in Oscillators Post by nrk1 on Apr 10th, 2016, 2:26am The result given below is incorrect. The AM-free signal that has the same shift in zero crossings as A0*cos(wot) + a*cos (wo+dw t + φ) is A0*cos(wot) + a/2*cos (wo+dw t + φ) + a/2*cos (wo-dw t + π - φ), for a≪A0. There is no factor of 1/2 for the carrier. Putting either of these signals through a limiter produces the same signal. The limiter's output around the fundamental will be proportional to the latter signal and depends on the limiting amplitude. iVenky wrote on Mar 29th, 2016, 8:42am:
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